A basis for a vector space Prove that for any non-zero linear function $f$ of a $n$-dimensional vector space $V$ there exists a basis ${e_1, ..., e_n}$ for the $V$ vector space, if 
$$f(x_1e_1+ ... + x_ne_n) = x_1$$ 
should be true for any $x_1, ..., x_n$
How to prove that? Thanks.
 A: So $V$ is an $n$-dimensional $K$ vector space. This fact you want to prove is true for any field $K$.
Since $f$ is nonzero, its range must be $K$. Therefore the nullspace of $f$ has dimension $n-1$ by the rank-nullity theorem.
Let $e_2,\ldots,e_n$ be a basis of this nullspace.
Then pick any $e_1$ such that $f(e_1)=1$.
It follows that $e_1,e_2,\ldots,e_n$ is a basis of $V$.
Indeed, if you take a linear combination of the latter and apply $f$, you'll see that the coefficient of $e_1$ is equal to $0$. Then you can use the linear independence of $e_2,\ldots,e_n$ to get the nullity of the other coefficients. So $e_1,\ldots, e_n$ are linearly independent. Given the dimension, they must form a basis of $V$.
Now
$$
f(x_1e_1+\ldots+x_ne_n)=\sum_{k=1}^n x_kf(e_k)=x_1
$$
as you wanted.
A: Here is a more tedious approach that is slightly constructive.
Let $v_k$ be a basis for $V$. Choose $w_1 \in  V$ such that $f(w_1) = 1$ ($f$ is non-zero, and linear, so we can do this, and $w_1$ is necessarily non-zero). Let $w_1 = \sum \alpha_k v_k$, and suppose for convenience that $\alpha_1 \neq 0$ (at least one of the $\alpha_k$ is non zero, so we can always do this).
Then it is straightforward to show that $w_1, v_2,...,v_n$ is a basis for $V$.
Now let $w_k = v_k -f(v_k) w_1$, for $k >1$, and note that $f(w_k) = 0$.
Then $w_1,...,w_n$ is a basis for $V$ with the required property. To  check that the $w_k$ are linearly independent we note that $\sum_k \alpha_k w_k = (\alpha_1-\sum_{k>1}\alpha_kf(v_k) )w_1 + \sum_{k>1} \alpha_k v_k$, hence if $\sum_k \alpha_k w_k = 0$, then $\alpha_k = 0$.
